Question

given: \\( \overline { k l } \parallel \overline { m j } \\), \\( \angle k j l \cong \angle m l j \\) prove: \\( \triangle j k l \cong \triangle l m j \\) proof: we know that \\( \overline { k l } \parallel \overline { m j } \\) and \\( \angle k j l \cong \angle m l j \\). because \\( \overline { k l } \parallel \overline { m j } \\), we know that \\( \angle k l j \cong \angle m j l \\) by the select choice theorem. we also know that \\( \overline { j l } \cong \overline { j l } \\) by the select choice of congruence. from this, we know that \\( \triangle j k l \cong \triangle l m j \\) by the select choice congruence postulate.

Explanation:

Step1: Alternate interior angles

Since \( \overline{KL}\parallel\overline{MJ} \), by the Alternate - Interior Angles Theorem, \( \angle KLJ\cong\angle MJL \).

Step2: Reflexive property of congruence

For any segment \( \overline{JL} \), \( \overline{JL}\cong\overline{JL} \) by the Reflexive Property of Congruence.

Step3: ASA congruence

We have \( \angle KJL\cong\angle MLJ \) (given), \( \overline{JL}\cong\overline{JL} \) (reflexive property), and \( \angle KLJ\cong\angle MJL \) (alternate - interior angles). So, by the Angle - Side - Angle (ASA) Congruence Postulate, \( \triangle JKL\cong\triangle LMJ \).

Answer:

Alternate - Interior Angles; Reflexive Property; ASA